235 



atmospheric resistance instead of H E sin A E H. If we take A' E = HE 

 and draw the perpendicular A' T, then A' T represents the height the body- 

 would rise (irrespective of gravity) in one second. JSTow A'T=AE sin AEH 

 = HEsin A EH. 



Again, Captain Hutton has unaccountably adopted a totally different 

 method to arrive at the vertical component of resistance in the case of the 

 wings, and has resolved the force represented by HE into (1) HK non- 

 effective (2) L E, resisting gravity, and (3) K L retarding the motion of the 

 bird. He has thus arrived at the strange conclusion that at one angle of 

 inclination (and for the body of the bird) the upward pressure is in proportion to 



. . /sin AEHv 

 the tangent of the angle, i.e., to the ratio of the sine to the cosme ( . „ TT ) 



& ° v cosAEH ; 



and that at another angle (for the wings) it is in proportion to the product of the 

 sine and cosine of the angle of inclination (sin C E H. cos C E H). The error 

 lies here. On his own assumption H E is the absolute velocity in one second, 

 therefore the retarding force has been overcome in addition to the production 

 of so much motion. The whole force exerted by the bird is, in fact, HE + R 

 ■where R : H E : : K L : K E, .-. it is not L E but K E ( = H E sin C E H) 

 which represents the vertical component of the force actually at work. Instead 

 therefore of H A and L E as measures of the upward pressures on the body and 

 wings respectively, we must take H E sin AEH and HE sin C E H. 



Captain Hutton goes on to say " the total amount the bird will rise per 

 second will be L E + H A feet." Introducing the corrections just made, this 

 amounts to saying that the upward pressure on the whole area of the bird 

 = H F (sin A E H + sin C E H). This is a grave error. 



Let P be the total pressure which supports the bird ; 



»p the average pressure on each square foot of sustaining surface ; 

 M the area of the lower surface of the body and tail ; 

 N the area of the under surface of the wings ; 



Then it is evident that P= M x H E sin A E H + N x H E sin C E H. 



P 



is ow by the assumed data p= -5 M = 2 and 1ST = 6 therefore 



HEsin AEH + 3. HEsin CEH /1X 



■■■P.= - j— • < • (!) 



Also p = 2 lbs. per square foot, and is assumed to represent the pressure of an 

 upward current of air having a velocity of 30 feet per second. From this we 

 obtain 



1 90 

 H F — : (2\ 



~ sin AEH + 3. sin CEH ' W 



This equation gives, when Z A E H = 0°, and C E H=15°, 



HE = T^£m = 15S < near W • • • • (3 > 



And for Z AEH=7°, and Z CEH = 22°, 



1 90 



HEz: — =96 (and a little more) . (4) 



.121869 + 3 x. 374607 V / W 



Captain Hutton closes this part of his calculations at this stage, and omits 

 to consider that as the angle of flight is increased, the sustaining surface is 

 reduced in the same proportion as the cosine of the Z of inclination to the 

 horizon. In passing from the conditions of equation (3) to those of (4), this 



I i 



